LeetCode:4Sum

Given an array S of n integers, are there elementsa, b, c, and d in S such that a +b + c + d = target?

Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,a ≤ b ≤ c ≤ d)

• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.    A solution set is:    (-1,  0, 0, 1)    (-2, -1, 1, 2)    (-2,  0, 0, 2)解题思路:    由于最终结果集中的四元组是非降序的，所以事先对序列进行排序，然后分别枚举c,d的位置.去重的做法与3Sum的做法类似，只是多了一重循环.时间复杂度为O(N^3).解题代码: 
class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target)    {        sort(num.begin(),num.end());        vector<vector<int> > res;        int n = num.size() , arr[4] ;        bool flag = false;        for(int d = n - 1 ; d > 2 ; --d)        {            if( d < n - 1 && num[d] == num[d+1])                continue;            for(int c = d - 1 ; c > 1 ; --c)            {                if( c < d - 1 && num[c] == num[c+1])                    continue;                int a = 0 , b = c - 1 ;                while( a < b)                {                    long long sum = num[a] + num[b] + num[c] + num[d] ;                    if(sum == target)                    {                        if(!flag)                        {                            arr[0] = num[a] , arr[1] = num[b] , arr[2] = num[c] , arr[3] = num[d];                            res.push_back(vector<int>(arr,arr+4));                            flag = true;                        }                        else if(arr[0] != num[a] || arr[1] != num[b] || arr[2] != num[c] || arr[3] != num[d])                        {                            arr[0] = num[a] , arr[1] = num[b] , arr[2] = num[c] , arr[3] = num[d];                            res.push_back(vector<int>(arr,arr+4));                        }                        ++a , --b ;                        continue;                    }                    sum > target ? --b : ++a ;                }            }        }        return res;    }};